First to draw an ace, alternating turns

Question

Two players alternate drawing a card from a full deck, each time returning it and reshuffling, with the first player drawing first. An ace appears on any draw with probability $\tfrac{1}{13}$ . Whoever draws an ace first wins.

What is the probability the first player wins?

Accepted Answer

Let P be the win probability of the player about to draw, with success probability p = 	frac{1}{13}. They win now with probability 	frac{1}{13}; otherwise (probability 	frac{12}{13}) the opponent draws next, winning with probability P, leaving 1 - P for the current player: P = 	frac{1}{13} + 	frac{12}{13}(1 - P) = 1 - 	frac{12}{13}P \;\Longrightarrow\; 	frac{25}{13}P = 1 \;\Longrightarrow\; P = \frac{13}{25} = 0.52. Takeaway. Using P = \frac{1}{2 - p} with p = 	frac{1}{13} gives \frac{13}{25}: for a rare event the first-mover edge is thin, barely above one-half.

First to draw an ace, alternating turns

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